The motion of an airplane is represented by a | vedclass

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(D) The distance covered is equal to the area under the velocity-time graph.For $t = 0$ to $t = 2 \ s$,the graph is a trapezoid with parallel sides $2

Vedclass · Accepted Answer

$12$

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(D) The distance covered is equal to the area under the velocity-time graph.For $t = 0$ to $t = 2 \ s$,the graph is a trapezoid with parallel sides $200 \ m/s$ and $400 \ m/s$,and height $2 \ s$.Area $1 = \frac{1}{2} 	imes (200 + 400) 	imes 2 = 600 \ m$.For $t = 2 \ s$ to $t = 30.5 \ s$,the velocity is constant at $400 \ m/s$.Time interval $\Delta t = 30.5 - 2 = 28.5 \ s$.Area $2 = 400 	imes 28.5 = 11400 \ m$.Total distance $= 600 + 11400 = 12000 \ m$.Converting to kilometers: $12000 \ m = 12 \ km$.

The motion of an airplane is represented by a velocity-time graph as shown below. The distance covered by the airplane in the first $30.5 \ s$ is . . . . . . $km$.

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